Download Thermodynamic Laws/Definition of Entropy Carnot Cycle

This is a review of Thermodynamics and Statistical Mechanics.
Thermodynamic Laws/Definition of Entropy
• 1st law of thermodynamics, the conservation of energy:
dU = dq − dw
∆ = Q − W,
(1)
where dq is heat entering the system and dw is work done by the system. Note the
convention: (+) for energy entering the system and (-) for energy leaving the system.
• 2nd law of thermodynamics, entropy: In any spontaneous transition, the entropy of
the universe increases.
There are many equivalent statements of the second law, such as: heat never flows from
cold to hot, or, there is no such thing as a perpetual motion machine.
In a reversible transformation, the entropy of the universe does not change. Note: this
does not mean that the entropy of some sub-universal system will not increase or decrease.
It is important to consider all parts of your system.
To find the change in entropy of a system between state A and state B, connect A and
B by a reversible path. Then
Z B
dq
.
(2)
S(B) − S(A) =
A T
Using Equation 2, we can illustrate the point about reversible transformations above:
Consider a reversible isothermal expansion of an ideal gas in contact with a thermal reservoir
at temperature T . For an ideal gas, U = 23 kT , for an isothermal expansion dU = 0, and
dq = P dV , so
Z
Z
Q
dq
P dV
=
= ∆S =
.
(3)
T
T
T
P =
nRT
,
V
so
Z
P
dV =
T
Z
VB
VA
VB
nR
dV = nR ln
> 0,
V
VA
but heat leaves the reservoir, thus ∆Sres =
−Q
T
(4)
and ∆Suniverse = 0.
Carnot Cycle
The Carnot Cycle is usually discussed with an ideal gas as the working substance, but in
reality any thermodynamic system (a paramagnet, an electrochemical cell, etc.) can be used.
A Carnot Cycle is a cycle involving two reversible isothermal transitions and two reversible
adiabatic transitions. If the working substance is an ideal gas, the P − V diagram of the
cycle looks like Fig. 1, and the S − T diagram looks like Fig. 2,
The efficiency of the Carnot Cycle is given by
η=
W
Q2 − Q1
T2 − T1
T1
=
=
=1− ,
Q2
Q2
T2
T2
1
(5)
Figure 1: Carnot Cycle: P − V diagram
Figure 2: Carnot Cycle: S − T diagram
which is work done by the system divided by the heat absorbed by the system. For a “Carnot
Refrigerator,” run the cycle backwards:
ηref =
Q1
Q1
=
=
W
Q2 − Q1
Q2
Q1
1
=
−1
T2
T1
1
T1
=
,
T2 − T1
−1
(6)
heat extracted from the system divided by work required to do so. Note that as T1 → T2 ,
ηref → ∞. In general, the efficiency is
η=
what you get out
.
what you need to put in
Typical Phase Diagrams
Figs. 3 and 4 show the phase diagrams for a single component system.
2
(7)
Figure 3: P − v diagram: Pressure vs. specific volume
Thermodynamic Potentials
Let’s begin with Eq. 1:
dU = dq − dw.
(8)
If we connect states by reversible processes we get T dS = dq and dW = P dV for a gas
system. So,
dU = T dS − P dV,
(9)
and setting one of ∂s or ∂v equal to zero yields,
∂U
∂U
= T,
= −P.
∂S V
∂V S
(10)
Define
∂F
∂T V
F ≡ U − T S,
dF = −SdT − P dV,
|
{z
}
Helmholtz free energy
Define
H ≡ U + P V, dH = T dS + V dρ,
|
{z
}
enthalpy
∂H
∂S ρ
= T,
=
∂F
∂V
∂H
∂ρ
T
= −S.
= V.
S
(11)
(12)
Define
G ≡ H − T S,
dG = −SdT + V dρ,
|
{z
}
Gibbs’ free energy
3
∂G
∂T ρ
= −S,
∂G
∂ρ
=V
T
(13)
Figure 4: P − T diagram: Pressure vs. Temperature. The dotted line characterizes the
anamolous behavior of water (and any substance that expands when frozen).
The differential form tells us the natural variables to use with each function. A, for example
(Fig. 11), is best expressed in terms of T and V : A(T, V ).
In physics we often do problems at constant (V, T ) and work with A. Chemists are prone
to a constant (T, ρ) and work with G. Note that we can get Maxwell’s relations from the
fact that each of these is an exact differential. For example,
∂F
∂F
dT +
dV,
(14)
dF =
∂T V
∂V T
thus
∂F
∂T V
but
so
= −S and
∂F
∂V
T
∂2F
∂2F
=
,
∂T ∂V
∂V ∂T
∂S
∂ρ
=
,
∂V T
∂T V
= −ρ,
(15)
(16)
(17)
which is one of Maxwell’s relations. Note from Dr. Collins: “I have not seen Maxwell’s
relations on any GRE example test (yet!)”.
Also, all of the above assume N is fixed. If N can vary each gets another term and we
must worry about chemical potentials:
∂U
dU = T dS − P dV + µdN where ∂N
= µ = chemical potential
(18)
S,V
The rest of the potentials also get this term, and we get a new potential, the grand potential
- Φ:
Φ = A − µN where dΦ = −SdT − ρdV − N dU
(19)
4
Heat Capacity
Heat capacity is specific heat if you divide by volume or mass
∆Q
holding volume constant
∆T ∂U
=
where U is internal energy
∂T V
∆U
cP =
holding pressure constant
∆T
cV =
(20)
(21)
(22)
cP > cV because the volume expands at constant pressure and the system does work which
means it stores less energy. Note that for non P V T systems analogous heat capacities can
be defined. In a magnetic system, P → H and V → M and we can define cH and cM .
A statistical approach can be used to obtain the classical heat capacity of common systems. If the Hamiltonian of a particle in the system can be written as
H=
N1
X
Aj Pj2
j
+
N2
X
Bj Q2j
(23)
j
where Pj and Qj are generalized momenta of coordinates, then
1
U = N kT f Equipartition Theorem
2
(24)
where N is the total number of particles and f = N1 + N2 , the number of quadratic coordinates and momenta. Thus,
1
cV = f N k
(25)
2
and the specific heat is 21 f k. You get 12 k per degree of freedom (momentum of coordinate
degrees of freedom count equally). For this expression to work, the temperature must be
large compared to the quantum of energy associated with a particular motion. For example,
if a diatomic molecule has a vibrational frequency ω, then the quantum of energy is ~ω
(E = (n + 12 )~ω). So kT hω to get N kT contributions to U ( 12 N λT for x2 and 12 N kT for
ρ2 ).
Consider the example of a diatomic molecule like N2 . See Fig. 5. We can write the
Hamiltonian as
1
1
1
1
1
1
1
(26)
H = M vx2cm + M vy2cm + M vz2cm + µṙ2 + kr2 + Iωz2 + Iωx2
2
2
2
2
2
2
2
Note that Iy = 0 for quantum mechanical reasons, so there is no rotational term about the
y-axis. We immediately know that at high temperatures
H = 72 kT N, cV = 72 N k, and cV =
cV
N
= 72 k.
(27)
One final thing: For and ideal gas, cV = 32 N k = 32 nR where N is the number of molecules,
n is the number
k is Boltzman’s constant, and R is the universal gas constant.
of moles,
∂H
5
Also cP = ∂T P = 2 N k. In general, for gases with ideal-like behavior, but internal degrees
of freedom (rotations, stretches), cP = cV + N k.
5
Figure 5: Diatomic molecule
Statistical Mechanics of systems at constant T and V
The probability a state of the system is occupied is proportional to e−Es /kT where E is the
energy of the state. The expectation value of the energy (which gives us U , the internal
energy, is
P
−Es /kT
s Es e
P
−Es /kT
se
If the system is made up of identical particles or a set of identical smaller systems (a bunch
of harmonic oscillators), we can just find the expected energy of the particle of a smaller
system, and multiply by N.
A couple of examples. Consider the two level system, a bunch of particles can have
energy O or E. The expectation value of the energy of N particles is
N Oe−O/kT + Ee−E/kT
< E >=
e−O/kT + e−E/kT
U=
N Ee−E/kT
1+e−E/kT
; as T → ∞, U → N E2 .
The average particle energy is halfway between the ground state and excited state.
If
U
E
> ,
N
2
then
e−E/kT
1
> ,
−E/kT
1+e
2
and
1
1
> −→ eE/kT < 1 and T < 0
E/kT
e
+1
2
6
(28)
(29)
(30)
Figure 6: number microstates vs. U
The two level problem is a classic problem which leads to negative temperatures. To see
why, let’s look at it a different way. Let’s plot the number of configurations of the system
that will give a particular energy. If the total energy is 0, there is only one configuration,
and in the ground state. Same for U = N E so the plot looks like Fig. 6. Here, S = k ln(Ω)
∂S
= T1 , where Ω is the multiplicity.
and dU = T dS − P dV , so ∂U
Refer to Fig. 6. When S is increasing, T is positive. When S is decreasing, T is negative.
This a characteristic of any system with a maximum total energy: the plot of the number of
microstates vs. energy will have a maximum, and thus negative temperatures.
Einstein model of a solid
The Einstein model is simply a bunch of atoms held together by springs. See Fig 7. We
approximate it as N harmonic oscillators with angular frequency ω. They are identical, so
we can just consider one.
P
n
hEi =
n+
P
n
We immediately see a trick: define β =
X
n
1
n+
2
1
.
kT
n+ 12
~ωe−(
1
~ωe−(n+ 2 )~ω/kT
1
e−(n+ 2 )~ω/kT
1
2
(31)
Then,
)~ωβ
∂
=−
∂β
7
!
X
n
n+ 12
e−(
)~ωβ
.
(32)
Figure 7: Einstein’s model of a solid
This is the same sum as in the denominator of Eq. 31, so let’s focus on it:
X
X
n
1
e−(n+ 2 )~ωβ = e−~ωβ/2
e~ωβ
| {z }
n
n
(33)
=x
= e−~ωβ/2
∞
X
xn
(34)
n=0
= e−~ωβ/2
1
1−x
(35)
e−~ωβ/2
1 − e−~ωβ
1
= ~ωβ/2
e
− e−~ωβ/2
1
1
=
2 sinh ~ωβ
2
(36)
=
and
∂
−1
1
−→
cosh
∂β
2 sinh ~ωβ 2
2
~ωβ
2
(37)
(38)
~ω
2
(39)
So,
hEi =
1
1
2 sinh( ~ωβ )
2
~ω
=
2
cosh
~ωβ
2
~ω
2
=
1
1
2 sinh( ~ωβ )
2
e~ωβ/2 + e−~ωβ/2
e~ωβ/2 − e−~ωβ/2
8
~ω
1
2 tanh ~ωβ
2
(40)
as T → ∞ and β → 0.
(41)
We can use series expansions:
and e−~ωβ/2 ≈ 1 −
~ω
2
1
hEi ≈
= = kT.
2 ~ωβ
β
e~ωβ/2 ≈ 1 +
~ω
β
2
~ω
β,
2
(42)
Lo and behold, we recover the equipartition result! 12 kT per degree of freedom and 2 degrees
of freedom: 12 mv 2 + 21 kx2 .
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